Q. 35 - Calculus

Question

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer. 

   ${\int }_0^1 \frac{1}{1+x^2}dx$

Step-by-Step Solution

Answer

Ans:  The exact value of, ${\int }_0^1 \frac{1}{1+x^2}dx =\frac{\mathrm{\pi }}{4}$.

Step 1. Given information.

given,

        ${\int }_0^1 \frac{1}{1+x^2}dx$

Step 2. The objective is to determine the exact value of the definite integral.

The exact value is calculated as shown below, 

      $\begin{array}{r}{\int }_0^1 \frac{1}{1+x^2}dx\\ ={\left[{\tan }^{-1}x\right]}_0^1\\ ={\tan }^{-1}\left(1\right)-{\tan }^{-1}0\\ =\frac{\pi }{4}\end{array}$

Therefore, the exact value is $\frac{\mathrm{\pi }}{4}$.

Step 3. Check

The required graph is,